What is the integral of 3-4sqrt(x) ?

The integral of 3-4sqrt(x) is 3x-8/3 x^{3/2}+C

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integral of 3-4sqrt(x)

\int 3 - 4 x d x

3 x - \frac{8}{3} x^{\frac{3}{2}} + C

Solution steps

\int 3 - 4 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 d x - \int 4 x d x

\int 3 d x = 3 x

\int 4 x d x = \frac{8}{3} x^{\frac{3}{2}}

= 3 x - \frac{8}{3} x^{\frac{3}{2}}

Add a constant to the solution

= 3 x - \frac{8}{3} x^{\frac{3}{2}} + C

\int \frac{1}{- x ^{2} - 16 x} d x

\int x 0.04 x^{\frac{3}{4}} + 4 d x

\frac{3}{2} x

\int x \cdot 3 x + 1 d x

\int y^{3} 1 + 9 y^{4} d y

What is the integral of 3-4sqrt(x) ?
The integral of 3-4sqrt(x) is 3x-8/3 x^{3/2}+C